# Are decompositions of a random variable into a sum of two IID random variables unique?

Let $$Z$$ be a real-valued random variable, and suppose that $$Z = X_1 + X_2$$ where $$X_1$$ and $$X_2$$ are i.i.d. random variables. Suppose further that $$Z = Y_1 + Y_2$$ where $$Y_1$$ and $$Y_2$$ i.i.d. random variables. Does it follow that $$Y_1$$ and $$X_1$$ are identically distributed?

I've looked into indecomposable distributions and infinitely divisible distributions, but could not find a result/example immediately answering the above question.

The answer is no in general.

In the language of characteristic functions, you have $$\phi_Z(t)=\phi^2_X(t)=\phi_Y^2(t)$$. Taking the square root implies $$\phi_Y(t)=\pm\phi_X(t)$$ for every $$t$$, in particular with $$\phi_X(t)=\phi_Y(t)$$ for $$t$$ in a neighborhood of 0 (since $$\phi(0)=1$$ for any characteristic function and by uniform continuity).

Thus, this is true if you require that $$\phi_X$$ and $$\phi_Y$$ are analytic, as they would necessarily agree on an interval, and therefore everywhere.

Otherwise it is not true, meaning that $$|\phi_X(t)|=|\phi_Y(t)|$$, such that there are $$t$$ where $$\phi_X(t)\neq \phi_Y(t)$$. A classic example of this is as follows:

Let $$P(X=x)=\frac{2}{\pi^2(2k-1)^2}$$ whenever $$x=\pm (2k-1)\pi$$ for $$k=1,2\cdots$$ , $$P(X=0)=1/2$$, and $$P(X=x)=0$$ otherwise.

Let $$P(Y=y)=\frac{4}{\pi^2(2k-1)^2}$$ whenever $$y=\pm (2k-1)\pi/2$$ and $$k=1,2,\cdots$$ and $$P(Y=y)=0$$ otherwise.

Both these give rise to "triangle-wave" characteristic functions that are periodic outside the interval specified:

$$\phi_X(t)=1-|t|, t\in[-1,1], \mbox{ periodic otherwise}$$ $$\phi_Y(t)=1-|t|, t\in[-2,2], \mbox{ periodic otherwise},$$

which look like: